Solve for $x$ : $ 3|x + 6| + 5 = 1|x + 6| + 2 $
Subtract $ {1|x + 6|} $ from both sides: $ \begin{eqnarray} 3|x + 6| + 5 &=& 1|x + 6| + 2 \\ \\ { - 1|x + 6|} && { - 1|x + 6|} \\ \\ 2|x + 6| + 5 &=& 2 \end{eqnarray} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} 2|x + 6| + 5 &=& 2 \\ \\ { - 5} &=& { - 5} \\ \\ 2|x + 6| &=& -3 \end{eqnarray} $ Divide both sides by ${2}$ $ \dfrac{2|x + 6|} {{2}} = \dfrac{-3} {{2}} $ Simplify: $ |x + 6| = -\dfrac{3}{2}$ The absolute value cannot be negative. Therefore, there is no solution.